Rank of Matrix A and Augmented Matrix (A|B)
Two systems may look almost identical, yet one has infinite solutions and the other has no solution at all. The answer is hidden in comparing rank(A) and rank(A|B).
What Is Matrix A and Augmented Matrix (A|B)?
A system of linear equations can be written as:
AX = B
- Matrix A is the coefficient matrix.
- Matrix (A|B) is the augmented matrix formed by appending column B to A.
What Is Rank of a Matrix?
The rank of a matrix is the maximum number of linearly independent rows (or columns). In simple terms, rank tells us how much independent information the matrix contains.
Many students find only rank(A) and forget to compare it with rank(A|B). This single mistake changes a correct method into a wrong exam answer.
Why Do We Compare rank(A) and rank(A|B)?
The comparison of rank(A) and rank(A|B) decides whether a system of linear equations is:
- Consistent with a unique solution
- Consistent with infinitely many solutions
- Inconsistent (no solution)
You do not need to solve the system completely.
Just compare rank(A) and rank(A|B).
Conditions Based on Rank
| Condition | Nature of Solution |
|---|---|
| rank(A) = rank(A|B) = n(number of variables) | Unique solution |
| rank(A) = rank(A|B) < n | Infinitely many solutions |
| rank(A) ≠ rank(A|B) | No solution (Inconsistent) |
What Is Row Echelon Form (REF)?
To find the rank of a matrix, we first convert it into row echelon form using elementary row operations. The rank of a matrix is equal to the number of non-zero rows in its row echelon form.
Definition Row Echelon Form(REF)
A matrix is said to be in row echelon form if it satisfies the following conditions:
- All non-zero rows are above any rows of all zeros.
- The first non-zero entry (leading entry) of each non-zero row is to the right of the leading entry of the row above it.
- All entries below each leading entry are zero.
Why Row Echelon Form (REF) Is Important
Row echelon form helps us:
- Determine the rank of a matrix easily.
- Check whether a system of linear equations is consistent or inconsistent.
- Identify free variables in case of infinite solutions.
Elementary Row Operations Used
The following row operations do not change the rank of a matrix:
- \(R_i \leftrightarrow R_j\) (Interchange two rows)
- \(R_i \rightarrow kR_i,\; k \neq 0\) (Multiply a row by a non-zero constant)
- \(R_i \rightarrow R_i + kR_j\) (Add a multiple of one row to another)
Rank from Row Echelon Form(REF)
After converting a matrix to row echelon form:
- The rank is the number of non-zero rows.
- A row of the form \([0\;0\;0]\) does not contribute to rank.
- A row of the form \([0\;0\;0\;|\;c]\) with \(c \neq 0\) indicates an inconsistent system.
Solved Example: Convert a Matrix into Row Echelon Form
Find the row echelon form of the matrix:
\[ A = \begin{bmatrix} 1 & 2 & 3\\ 2 & 4 & 6\\ 3 & 5 & 7 \end{bmatrix} \]
Step 1: Make zeros below the first pivot
The first pivot is the entry 1 in the first row and first column. We eliminate the entries below it.
Apply the row operations:
\[ R_2 \rightarrow R_2 - 2R_1, \qquad R_3 \rightarrow R_3 - 3R_1 \]
After applying these operations, we get
\[ \sim \begin{bmatrix} 1 & 2 & 3\\ 0 & 0 & 0\\ 0 & -1 & -2 \end{bmatrix} \]
Step 2: Arrange non-zero rows above zero rows
Interchange the second and third rows to move the non-zero row upward.
\[ R_2 \leftrightarrow R_3 \]
We obtain
\[ \sim \begin{bmatrix} 1 & 2 & 3\\ 0 & -1 & -2\\ 0 & 0 & 0 \end{bmatrix} \]
Step 3: Final Row Echelon Form
The matrix now satisfies all conditions of row echelon form:
- All zero rows are at the bottom.
- Each leading entry is to the right of the leading entry in the row above.
- All entries below each pivot are zero.
\[ \text{REF of } A = \begin{bmatrix} 1 & 2 & 3\\ 0 & -1 & -2\\ 0 & 0 & 0 \end{bmatrix} \]
\[ \operatorname{rank}(A) = 2 \]
Key Exam Note
Always reduce both the coefficient matrix \(A\) and the augmented matrix \((A|B)\) to row echelon form before comparing their ranks.
Example 1: Looks Different but Has Infinite Solutions
The coefficient matrix is
\[ A = \begin{bmatrix} 1 & 1\\ 2 & 2 \end{bmatrix} \]
Apply the row operation:
\[ R_2 \rightarrow R_2 - 2R_1 \]
After applying the operation, we get
\[ A \sim \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix} \]
There is only one non-zero row.
\[ \operatorname{rank}(A)=1 \]
Finding the Rank of the Augmented Matrix (A|B)
The augmented matrix is
\[ (A|B)= \begin{bmatrix} 1 & 1 & | & 2\\ 2 & 2 & | & 4 \end{bmatrix} \]
Apply the same row operation:
\[ R_2 \rightarrow R_2 - 2R_1 \]
After row reduction, we obtain
\[ (A|B)\sim \begin{bmatrix} 1 & 1 & | & 2\\ 0 & 0 & | & 0 \end{bmatrix} \]
Again, there is only one non-zero row.
\[ \operatorname{rank}(A|B)=1 \]
\[ \operatorname{rank}(A) = \operatorname{rank}(A|B) = 1. \] Since the number of variables is \(n = 2\), and \[ \operatorname{rank}(A) = \operatorname{rank}(A|B) < n, \] the system is consistent with infinitely many solutions.Same ratio of coefficients and constants leads to dependent equations, not inconsistency.
Example 2: Looks Same but Has No Solution
We apply the elementary row operation:
\[ R_2 \rightarrow R_2 - R_1 \]
\[ (A|B)\sim \begin{bmatrix} 1 & 1 & | & 2\\ 0 & 0 & | & 1 \end{bmatrix} \]
The second row represents the equation
Hence, the augmented matrix has two non-zero rows.
\[ \operatorname{rank}(A|B)=2 \]
\[ \operatorname{rank}(A) = 1, \quad \operatorname{rank}(A|B) = 2. \] Since \[ \operatorname{rank}(A) \neq \operatorname{rank}(A|B), \] the system is \(\textbf{inconsistent and has no solution}\).Three-Variable Systems Using Rank Method
Now we consider systems of linear equations involving three variables. The comparison of rank(A) and rank(A|B) decides the nature of solutions.
Case 1: Unique Solution
This case occurs when:
\[ \operatorname{rank}(A) = \operatorname{rank}(A|B) = 3 \]
Example (Three Variables – Unique Solution):
\[ \begin{aligned} x + y + z &= 6,\\ 2x + y + z &= 7,\\ x + 2y + 3z &= 14. \end{aligned} \]
Coefficient matrix:
\[ A = \begin{bmatrix} 1 & 1 & 1\\ 2 & 1 & 1\\ 1 & 2 & 3 \end{bmatrix} \]
Steps to Find the Rank of the Augmented Matrix (A|B)
The augmented matrix is
\[ (A|B)= \begin{bmatrix} 1 & 1 & 1 & | & 6\\ 2 & 1 & 1 & | & 7\\ 1 & 2 & 3 & | & 14 \end{bmatrix} \]
Step 1: Eliminate the first column entries below the pivot.
Apply the row operations:
\[ R_2 \rightarrow R_2 - 2R_1, \qquad R_3 \rightarrow R_3 - R_1 \]
We obtain
\[ \sim \begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & -1 & -1 & | & -5\\ 0 & 1 & 2 & | & 8 \end{bmatrix} \]
Step 2: Eliminate the second column entry below the pivot.
Apply the row operation:
\[ R_3 \rightarrow R_3 + R_2 \]
This gives
\[ \sim \begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & -1 & -1 & | & -5\\ 0 & 0 & 1 & | & 3 \end{bmatrix} \]
Step 3: Observe the row-echelon form.
All three rows are non-zero.
\[ \operatorname{rank}(A|B)=3 \]
Since the coefficient matrix undergoes the same row operations,
\[ \operatorname{rank}(A)=3 \]
\[ \operatorname{rank}(A)=\operatorname{rank}(A|B)=3=n \]
Hence, the system has a unique solution.
Case 2: Infinitely Many Solutions
This case occurs when:
\[ \operatorname{rank}(A) = \operatorname{rank}(A|B) < 3 \]
Example (Three Variables – Infinite Solutions):
\[ \begin{aligned} x + y + z &= 3,\\ 2x + 2y + 2z &= 6,\\ 3x + 3y + 3z &= 9. \end{aligned} \]
Coefficient matrix:
\[ A = \begin{bmatrix} 1 & 1 & 1\\ 2 & 2 & 2\\ 3 & 3 & 3 \end{bmatrix} \]
Augmented matrix:
\[ (A|B) = \begin{bmatrix} 1 & 1 & 1 & | & 3\\ 2 & 2 & 2 & | & 6\\ 3 & 3 & 3 & | & 9 \end{bmatrix} \]
Steps to Find the Rank of the Augmented Matrix (A|B)
The augmented matrix is
\[ (A|B)= \begin{bmatrix} 1 & 1 & 1 & | & 3\\ 2 & 2 & 2 & | & 6\\ 3 & 3 & 3 & | & 9 \end{bmatrix} \]
Step 1: Eliminate the first column entries below the pivot.
Apply the row operations:
\[ R_2 \rightarrow R_2 - 2R_1, \qquad R_3 \rightarrow R_3 - 3R_1 \]
We obtain
\[ \sim \begin{bmatrix} 1 & 1 & 1 & | & 3\\ 0 & 0 & 0 & | & 0\\ 0 & 0 & 0 & | & 0 \end{bmatrix} \]
Step 2: Observe the row-echelon form.
Only one row is non-zero.
\[ \operatorname{rank}(A|B)=1 \]
Since the coefficient matrix undergoes the same row operations,
\[ \operatorname{rank}(A)=1 \]
\[ \operatorname{rank}(A)=\operatorname{rank}(A|B)=1<3 \]
Hence, the system is consistent with infinitely many solutions.
Case 3: No Solution (Inconsistent System)
This case occurs when:
\[ \operatorname{rank}(A) \neq \operatorname{rank}(A|B) \]
Example (Three Variables – No Solution):
\[ \begin{aligned} x + y + z &= 3,\\ x + y + z &= 4,\\ 2x + 2y + 2z &= 6. \end{aligned} \]
Coefficient matrix:
\[ A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 2 & 2 & 2 \end{bmatrix} \]
Augmented matrix:
\[ (A|B) = \begin{bmatrix} 1 & 1 & 1 & | & 3\\ 1 & 1 & 1 & | & 4\\ 2 & 2 & 2 & | & 6 \end{bmatrix} \]
Steps to Find the Rank of the Augmented Matrix (A|B)
The augmented matrix is
\[ (A|B)= \begin{bmatrix} 1 & 1 & 1 & | & 3\\ 1 & 1 & 1 & | & 4\\ 2 & 2 & 2 & | & 6 \end{bmatrix} \]
Step 1: Eliminate the first column entries below the pivot.
Apply the row operations:
\[ R_2 \rightarrow R_2 - R_1, \qquad R_3 \rightarrow R_3 - 2R_1 \]
We obtain
\[ \sim \begin{bmatrix} 1 & 1 & 1 & | & 3\\ 0 & 0 & 0 & | & 1\\ 0 & 0 & 0 & | & 0 \end{bmatrix} \]
\[ \operatorname{rank}(A|B)=2 \]
Now consider the coefficient matrix.
After applying the same row operations to \(A\), we get
\[ A \sim \begin{bmatrix} 1 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \]
\[ \operatorname{rank}(A)=1 \]
\[ \operatorname{rank}(A) \neq \operatorname{rank}(A|B) \]
Hence, the system is inconsistent and has no solution.
Master Rule (For Exams)
Why This System Has No Solution (Geometric View)
The equations represent two parallel lines with different intercepts. Parallel lines never intersect, so the system has no solution.
Frequently Asked Questions
No. The augmented matrix contains all columns of A, so its rank is always greater than or equal to rank(A).
Yes. Rank method is faster and works for any number of variables.
Yes. It works for systems with any number of variables.
Because the augmented matrix represents an extra equation that contradicts the original system, leading to an impossible equation such as \(0 = 1\).
No. If the ranks are equal, the system is always consistent and has at least one solution.
It indicates the presence of free variables, which leads to infinitely many solutions when the system is consistent.
Yes. Rank is determined by reducing the matrix to row-echelon or reduced row-echelon form.
No. Elementary row operations do not change the rank of a matrix.
Yes. Rank method works for rectangular matrices as well.
The maximum rank of a matrix is the smaller of the number of rows and the number of columns.
Because dependent equations do not provide new information and become zero rows after row reduction.
Yes. Rank method is commonly asked in CBSE, state boards, JEE, NET, and GATE examinations.
No. Rank method applies to both homogeneous and non-homogeneous systems of linear equations.
In this case, all equations reduce to \(0 = 0\), and the system has infinitely many solutions.
Always write the rank condition explicitly. Even if the final answer is wrong, partial marks are awarded.
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